-16t^2+20=0

Solution for -16t^2+20=0 equation:

-16t^2+20=0

a = -16; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-16)·20
Δ = 1280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*-16}=\frac{0-16\sqrt{5}}{-32}
=-\frac{16\sqrt{5}}{-32}
=-\frac{\sqrt{5}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*-16}=\frac{0+16\sqrt{5}}{-32}
=\frac{16\sqrt{5}}{-32}
=\frac{\sqrt{5}}{-2}
$